I think this is what you are looking for.
NeutrinoTrident[MZprime_, s24squad_] := (s24squad/MZprime)^2
UpperBound = (1/370)^2;
ContourPlot[
NeutrinoTrident[MZprime, s24squad], {MZprime, 100, 1000}, {s24squad, 0, 1},
ScalingFunctions -> {"Log", "Log"},
FrameLabel ->
{Style[Subscript[M, Derivative[1][Z]][GeV], FontSize -> 16],
Style[Row @ {sin^2, Subscript[θ, 24]}, FontSize -> 16]},
BaseStyle -> {FontFamily -> "Times", FontSize -> 14},
RegionFunction -> (NeutrinoTrident[#1, #2] <= UpperBound &),
PlotRange -> {{100 - 10^-10, 1000}, {10^-3, 1}},
Contours -> {UpperBound},
ContourShading -> {LightBlue}]

Update
After reading @kglr's answer here, I realized that this answer also needed to give PlotRange
the 3rd element All
. If I had done that, you probably wouldn't have needed to ask your follow-up question.
Here is the corrected version of my the above plot:
ContourPlot[
NeutrinoTrident[MZprime, s24squad], {MZprime, 100, 1000}, {s24squad, 0, 1},
ScalingFunctions -> {"Log", "Log"},
FrameLabel ->
{Style[Subscript[M, Derivative[1][Z]][GeV], FontSize -> 16],
Style[Row@{sin^2, Subscript[\[Theta], 24]}, FontSize -> 16]},
BaseStyle -> {FontFamily -> "Times", FontSize -> 14},
RegionFunction -> (NeutrinoTrident[#1, #2] <= UpperBound &),
PlotRange -> {{100 - 10^-10, 1000}, {10^-3, 1}, All},
ContourShading -> {LightBlue}]

In the case where you want have the contour at UpperBound
to be visible, then the corrected plot would be:
ContourPlot[
NeutrinoTrident[MZprime, s24squad], {MZprime, 100, 1000}, {s24squad, 0, 1},
ScalingFunctions -> {"Log", "Log"},
FrameLabel ->
{Style[Subscript[M, Derivative[1][Z]][GeV], FontSize -> 16],
Style[Row@{sin^2, Subscript[\[Theta], 24]}, FontSize -> 16]},
BaseStyle -> {FontFamily -> "Times", FontSize -> 14},
RegionFunction -> (NeutrinoTrident[#1, #2] <= UpperBound &),
PlotRange -> {{100 - 10^-10, 1000}, {10^-3, 1}, All},
Contours -> {.995 UpperBound},
ContourStyle -> {Thick},
ContourShading -> {LightBlue}]

RegionPlot
instead ofContourPlot
or use the optionContours->{}
inside ContourPlot $\endgroup$ – Ulrich Neumann Mar 5 '19 at 16:00RegionPlot
but there were lots of difficulties when I turned it into LogLog scale. So I thought drawing the plot in theContourPlot
is easier than turning the plot into LogLog scale inRegionPlot
. $\endgroup$ – lhcQFT Mar 5 '19 at 16:05